Double Integral over region?

Question

let | = integral of

| | over region R of -xydA

where R is the region in the first quadrant bounded by x^2 +y^2 = 16, y=3x, y=0

Answer 1

Use polar coordinates. Then R is the region bounded by 0 <= r <= 4, and 0 <= t <= arctan(3).

You then have |0,4 |0,arctan(3) -r^3 cos(t)sin(t) dt dr
= |0,4 -r^3 dr * |0,arctan(3) cos(t) sin(t) dt
= -r^4/4 |0,4 * (1/2)sin^2(t) |0,arctan(3)
= -64 * (1/2)(0.9)
= -28.8